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绵阳市高中2020级第二次诊断性考试文科数学参考答案及评分意见一、选择题：本大题共12小题，每小题5分，共60分．DDCAABCDBACA二、填空题：本大题共4小题，每小题5分，共20分．13．214．1−15．3−16．[1，3)三、解答题：本大题共6小题，共70分．17．解：（1）由

23(cos)sinbaCaC−=，及正弦定理可得，3sin3sincossinsinBACaAC−=，···············································2分∵3sin3sin()3sincos3cossinBACACAC=+=+···

···································4分∴3cossinsinsinACaAC=，·······························································6分即sin3cosaAA=，且3A=，可

得3a=；············································8分（2）由2121)cos(−=−=−=bcAbcACBA，可得1cb=，······················10分由余弦定理2222cos4bc

abcA+=+=.··················································12分18．解：（1）由题意知，2nS=2na+na，①···············································1分当n=1时，21a=21

a+1a，则11a=；··························································2分当2≥n时，21nS−=21na−+1na−，②·····························

························3分①②相减可得，2an=2na−21na−+na−1na−，················································4分∴an+1na−=2na−21na−，则an

-1na−=1，∴数列na是以11a=为首项，1为公差的等差数列，··································5分所以，an=n(n∈N∗).·························

···············································6分（2）2()3nnnabn=，·····························································

··············7分设nnncab=，则1112232()(1)()()3333nnnnnnccnn−−−−−=−−=，························8分∴当3n时，10nncc−−，所以1nncc−，··········

·································9分当3n=时，10nncc−−=，所以1nncc−=，·············································10分当3n时，10nncc−−，所以1nncc−，··············

·······························11分则12345ccccc=，∴存在23或m=，使得对任意的，≤nnmmnNabab恒成立．·····················12分19．解：（1）因为0.92<0.99，根据统计学相关知识，2R越大，意味着

残差平方和521ˆ()iiyy=−越小，那么拟合效果越好，因此选择非线性回归方程②2ˆˆˆymxn=+进行拟合更加符合问题实际．·······························································4分（2）令2iiux=，则先求出线性回归

方程：ˆˆˆymun=+，································5分∵14916250.81.11.52.43.7111.955=，uy++++++++===，·····················7分2222221()(111)(411)(911)(1611)

(2511)niiuu=−=−+−+−+−+−=374，···········9分∴121()()45.1ˆ0.121374()niiiniiuuyymuu==−−==−，··········

······································10分由ˆ1.90.12111n=+，得ˆ0.5690.57n=，即ˆ0.120.57yu=+，································

··········································11分∴所求非线性回归方程为：2ˆ0.120.57yx=+．········································12分20．解：（1）设11()Bxy，，22()

Cxy，，直线BC的方程为：4xmy=+，其中1=mk，·······································1分联立224143xmyxy=++=，消x整理得：22(34)24360mymy+++=，··················

·2分所以：1222434myym−+=+，1223634yym=+，············································3分从而121212121222(6)(6)yyyykkxxmymy==++++

12212126()36yymyymyy=+++2222236134361444363434mmmmm+==−+++所以：12kk为定值14．···························································

···········5分（2）直线AB的方程为：)2(211++=xxyy，············································6分令4x=，得到66261111+=

+=myyxyyM，·················································7分同理：2266Nyymy=+．·····································

····································8分从而121266||||||66MNyyMNyymymy=−=−++122121236|||6()36|yymyymyy−=+++······································

···············9分又221212122124||()434myyyyyym−−=+−=+，212122144|6()36|34myymyym+++=+，····················································10分所以2||34MNm

=−，······································································11分因为：1[34]，mk=，所以||[3563]，MN，即线段MN长度的取值范围为[3563]，．··················

··························12分21．解：（1）解：(1)a=2时，2()ln32fxxxx=+−+，2231(21)(1)()xxxxfxxx−+−−==，·········································

·············2分由()0fx解得：x>1或102x；由()0fx解得：112x．················3分故f(x)在区间(1)，+，1(0)2，上单调递增，在区间1(1

)2，上单调递减．··········4分所以f(x)的极大值是13()ln224f=−，极小值是f(1)=0；······························5分（2）2(1)1(1)(1)()axaxaxx

fxxx−++−−==，且10≥x−，·········································6分①当≥1a时，10≥ax−，(1)(1)()0≥axxfxx−−=，故f(

x)在区间[1，2]上单调递增，所以min()()0fxha==，····························7分②当102≤a时，10ax−，(1)(1)()0≤axxfxx−−=，故f(x)在区间[1，2]上单调递减，所以min()()(

2)ln2102≥afxhaf===+−，显然()ha在区间1(0]2，上单调递增，故13()()ln224≤hah=−<0．·······················································

··············9分③当112a时，由()0fx解得：12≤xa；由()0fx解得：11≤xa．故f(x)在区间1(2]，a上单调递增，在区间1[1)，a上单调递减．此时min11

()()()ln22afxfhaaaa===−−，则222111(1)()0222≥ahaaaa−=−+=，故()ha在区间1(1)2，上单调递增，故h(a)<h(1)=0.·······································11分综

上：011()ln2102211ln1222，≥，≤，aahaaaaaa=+−−−，且h(a)的最大值是0.··························12分22．解：（1）①当B在线段AO上

时，由|OA|‧|OB|=4，则B（2，）或（2，23）；②当B不在线段AO上时，设B（ρ，θ），且满足|OA|‧|OB|=4，∴A4（，）+，·······················

·····················································1分又∵A在曲线l上，则44cos()sin()2+++=−，·······················

····3分∴2sin2cos=+，·····································································4分又∵3≤≤2+，即20≤≤.综上所述，曲线C的极坐标方程为：2sin2cos=+2（0≤≤

），或32()2=或==.··························5分（2）①若曲线C为：32()2=或==，此时P，Q重合，不符合题意；②设l1：=2（0≤≤），又l1与曲线C交于点P，联立2sin2cos，，

==+得：2sin2cosP=+，···································································6分又l1与曲线l交于点Q，联立sincos2，，=+=−得：2sincosQ−=+，·

·····································································7分又∵M是P，Q的中点，1sincos(0)2sincos2≤≤PQM+==+−+，·······

·······················8分令sincost+=，则2sin()4t=+，又∵20≤≤，则3444≤≤+，且12≤≤t，∴1(12)≤≤Mttt=−，且1Mtt=

−在12，上是增函数，·····················9分∴22221≤M−=，且当42+=时，即4=时等号成立．∴OM的最大值为22．·····························

······································10分23．解：（1）由()fx≤3的解集为[n，1]，可知，1是方程()fx=3的根，∴(1)f=3+|m+1|=3，则m=−1，·······························

································1分∴()fx=|2x+1|+|x−1|，①当x≤12−时，()fx=−3x≤3，即x≥−1，解得：−1≤x≤12−，···············

·2分②当112x时，()fx=x+2≤3，解得：112x−，································3分③当x≥1时，()fx=3x≤3，解得：x=1．···············································4分综上所述：(

)fx的解集为[−1，1]，所以m=−1，n=−1．·····························5分（2）由（1）可知m=−1，则1222ab+=.·······················································6分令12xa=，2

yb=，则12ax=，2by=，又a，b均为正数，则2xy+=（00，xy），由基本不等式得，22≥xyxy=+，······················································7分

∴1≤xy，当且仅当，x=y=1时等号成立．所以有11≥xy，当且仅当，x=y=1时等号成立．········································8分又22222244164(2

)ababxy+=+=+224482≥xyxy=(当且仅当，x=y时等号成立)．·········9分∴22168≥ab+成立，(当且仅当，122，ab==时等号成立)．····················10分